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Your friend decides to produce electrical power by turning a coil of \(1.00 \cdot 10^{5}\) circular loops of wire around an axis parallel to a diameter in the Earth's magnetic field, which has a local magnitude of \(0.300 \mathrm{G}\). The loops have a radius of \(25.0 \mathrm{~cm} .\) a) If your friend turns the coil at a frequency of \(150.0 \mathrm{~Hz}\) what peak current will flow in a resistor, \(R=1500 . \Omega\) connected to the coil? b) The average current flowing in the coil will be 0.7071 times the peak current. What will be the average power obtained from this device?

Short Answer

Expert verified
Step 1: Find the peak magnetic flux Φ_B_peak = (B * π * r^2) = (0.3 * 10^{-4} T * π * (0.25 m)^2) = 5.896 × 10^{-6} T m^2 Step 2: Calculate the peak EMF ω = 2 * π * f = 2 * π * 150 Hz = 942.48 rad/s ε_peak = N * Φ_B_peak * ω = (1.00 * 10^{5}) * (5.896 × 10^{-6} T m^2) * (942.48 rad/s) = 0.2779 V Step 3: Calculate the peak current I_peak = ε_peak / R = (0.2779 V) / (1500 Ω) = 1.852 × 10^{-4} A Step 4: Calculate the average power I_avg = 0.7071 * I_peak = 0.7071 * (1.852 × 10^{-4} A) = 1.309 × 10^{-4} A P_avg = I_avg^2 * R = (1.309 × 10^{-4} A)^2 * (1500 Ω) = 2.557 × 10^{-6} W The peak current flowing in the resistor is approximately 1.852 × 10^{-4} A, and the average power obtained from this device is approximately 2.557 × 10^{-6} W.

Step by step solution

01

Find the peak magnetic flux

To find the peak magnetic flux, we need to calculate the maximum possible magnetic flux through each loop. The magnetic flux (Φ_B) is given by: Φ_B = B * A * cos(θ) where: B - the magnetic field magnitude, A - the area of the loop, θ - the angle between the magnetic field and the normal to the loop surface. Since the loop is rotated parallel to the Earth's magnetic field diameter, at the maximum possible moment, the normal to the loop surface will be perpendicular to the magnetic field, and the angle θ will be 0 degrees. Therefore, cos(θ) is 1 at this moment. The area A of the loop can be calculated as: A = π * r^2 Using the given magnetic field magnitude (B = 0.3 G = 0.3 * 10^{-4} T) and the radius of the loops (r = 25 cm = 0.25 m), we get: Φ_B_peak = (B * π * r^2)
02

Calculate the peak EMF

Next, we will use Faraday's Law of electromagnetic induction to find the peak electromotive force (EMF) induced in the coil. Faraday's Law states that the induced EMF (ε) in a coil is equal to the negative rate of change of magnetic flux (Φ_B) with respect to time (t): ε = -N * d(Φ_B) / dt where N is the number of loops in the coil. Since the coil is turning at a frequency f, the angular velocity (ω) can be found as: ω = 2 * π * f The change of magnetic flux with respect to time can be determined as the product of the peak flux and the derivative of cos(ωt): d(Φ_B) / dt = Φ_B_peak * (-ω * sin(ωt)) The induced EMF at any time t is: ε(t) = N * (Φ_B_peak * ω * sin(ωt)) The peak EMF (ε_peak) occurs when sin(ωt) = 1: ε_peak = N * Φ_B_peak * ω Using the given values for N (1.00 * 10^{5}), f (150 Hz), and our previous calculation for Φ_B_peak, find the peak EMF value: ε_peak = N * Φ_B_peak * ω
03

Calculate the peak current

Now we will use Ohm's Law to find the peak current (I_peak) flowing in the resistor: I_peak = ε_peak / R Where R is the resistance (1500 Ω). Use the peak EMF value calculated in Step 2 to calculate the peak current.
04

Calculate the average power

The average current flowing in the coil (I_avg) is given as: I_avg = 0.7071 * I_peak The average power (P_avg) can be calculated as: P_avg = I_avg^2 * R Using the calculated values for the average current and resistance, find the average power obtained from this device.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Flux
Magnetic flux, denoted by \( \Phi_B \), is a measure of the magnetic field passing through a given area. It's like imagining lines of a magnetic field passing through a loop, providing insight into the magnetic influence over a surface. Mathematically, magnetic flux is calculated as \( \Phi_B = B \cdot A \cdot \cos(\theta) \), where:
  • \( B \) is the magnetic field strength.
  • \( A \) is the area the field lines pass through.
  • \( \theta \) is the angle between the field direction and normal to the surface.
In scenarios like rotating coils, understanding how magnetic flux changes is essential, especially because a perpendicular alignment (\( \theta = 0 \)) leads to the maximum flux, which occurs when the loop is optimally positioned regarding the magnetic field.
Faraday's Law
Faraday's Law of electromagnetic induction is a fundamental principle that describes how electric currents can be induced in a loop or coil by changing the magnetic flux through it. Roughly speaking, it connects the change in magnetic environment with current production. According to Faraday's Law, the induced electromotive force (EMF), \( \varepsilon \), is equal to the negative rate of change of the magnetic flux through the coil:\[ \varepsilon = -N \frac{d\Phi_B}{dt} \]where:
  • \( N \) is the number of loops.
  • \( \frac{d\Phi_B}{dt} \) represents the rate of change of magnetic flux.
The negative sign conveys Lenz's Law, which tells us the induced EMF works to oppose the change in flux, maintaining a sort of electromagnetic balance. By rotating a coil in a magnetic field, the flux changes and hence generates an EMF, leading to current if connected to a circuit.
Peak Current
The peak current refers to the maximum instantaneous value of current that can flow in a circuit. In the context of our rotating coil, we determine the peak current using the peak electromotive force (EMF) calculated from Faraday's Law. By applying Ohm's Law, the peak current, \( I_{\text{peak}} \), is determined using:\[ I_{\text{peak}} = \frac{\varepsilon_{\text{peak}}}{R} \]where:
  • \( \varepsilon_{\text{peak}} \) is the peak induced EMF.
  • \( R \) is the resistance in the circuit.
This measurement is crucial as it indicates the highest possible current level the system can handle without undergoing modifications, and can be linked to the device's performance during peak conditions.
Ohm's Law
Ohm's Law is a simple yet powerful law that defines the relationship between voltage, current, and resistance in an electrical circuit. This law is fundamental in understanding how circuits operate and is represented as:\[ V = I \cdot R \]where:
  • \( V \) is the voltage across a component.
  • \( I \) is the current flowing through the component.
  • \( R \) is the resistance the component offers.
In terms of determining current from voltage or vice versa, Ohm's Law helps in calculating the peak current when the peak EMF is known in scenarios involving induced EMFs. It's instrumental in deciding how variations in resistance or voltage will impact electrical current, ensuring circuits operate efficiently and safely.

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Most popular questions from this chapter

A solid metal disk of radius \(R\) is rotating around its center axis at a constant angular speed of \(\omega .\) The disk is in a uniform magnetic field of magnitude \(B\) that is oriented normal to the surface of the disk. Calculate the magnitude of the potential difference between the center of the disk and the outside edge.

A steel cylinder with radius \(2.5 \mathrm{~cm}\) and length \(10.0 \mathrm{~cm}\) rolls without slipping down a ramp that is inclined at \(15^{\circ}\) above the horizontal and has a length (along the ramp) of \(3.0 \mathrm{~m} .\) What is the induced potential difference between the ends of the cylinder at the bottom of the ramp, if the surface of the ramp points along magnetic north?

A solenoid with 200 turns and a cross-sectional area of \(60 \mathrm{~cm}^{2}\) has a magnetic field of \(0.60 \mathrm{~T}\) along its axis. If the field is confined within the solenoid and changes at a rate of \(0.20 \mathrm{~T} / \mathrm{s}\), the magnitude of the induced potential difference in the solenoid will be a) 0.0020 V. b) \(0.02 \mathrm{~V}\). c) 0.001 V. d) \(0.24 \mathrm{~V}\).

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A circular coil of wire with 20 turns and a radius of \(40.0 \mathrm{~cm}\) is laying flat on a horizontal table as shown in the figure. There is a uniform magnetic field extending over the entire table with a magnitude of \(5.00 \mathrm{~T}\) and directed to the north and downward, making an angle of \(25.8^{\circ}\) with the horizontal. What is the magnitude of the magnetic flux through the coil?

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